On Fri, Apr 15, 2011 at 10:59 AM, Sean Breheny wrote: > Right, but if you measure the emitter current and try to control it > to, say, 1 Amp, and the base needs 10mA drive current. Then, you will > only be pulling 990mA from the battery so you will be off by 1%. That > might be acceptable (or maybe not depending on the application). > What's worse is that the current gain of many BJTs at currents of > several amps may be very low, perhaps even as low as 10, so you could > be 10% off in your measurement in that case. One thing going in your > favor is that you are not operating the BJT as a "saturated switch" > (i.e., full on) so you will not be in BJT saturation and the current > gain will stay closer to the nominal value for that BJT. Also, you > could use a Darlington which should easily get you up in the gain of > >100 range even at high currents. > > Sean > > Unfortunately, I haven't had the time lately to read in depth on how BJTs work. I'm not in the EE program, so any EE I do, I have to do solely out of my spare time, which is quite sparse in the undergrad life science field. I didn't know the gain worked like that. I assumed 10 mA into the base =3D 1000 mA pulled by the collector. I tested it on LTSpice and that's what I saw as well. If I drive the base with 10 mA, why does the collector only pull 990 mA? Also, if I use a Darlington pair, should I use a small transistor as the first transistor and the large transistor for the second? Or should both of them be the same large transistor? To measure current, I'll just measure the voltage drop across the load resistor and adjust the base current according to that. --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .